In WPF, I am trying to create a settings page like the one in Visual Studio (Tools -> Options) (http://i.imgur.com/Be9cTPF.png)
I realize the selection menu is a treeview but I am unsure how to hide/show the controls on the right.
What is the best and proper way of doing this in WPF?
In my opinion the best way to do this is to divide the page into 2 grid columns. In your first column, you can place the treeview. The second column will consist of the data page. You can bind the visibility of this grid column when there is a selection in the tree view using a converter or so or display a default view there.
You can handle the selection changed event of the treeview and bind the selecteditem to a property and as per the selecteditem you can change the view on the second column.
Hope this helps.
Related
How add different controls in one column GridView?
I want it to be like this.
Without creating separate gridviews, for other controls.
As far as I see, each row has its own type and the same editor should edit cells in this row. For this scenario, I think it's better to use the Vertical Grid instead of the Grid View.
If you still wish to use the Grid View, use the GridView.CustomRowCellEdit and GridView.CustomRowCellEditForEditing events to assign the required editor to a grid cell.
I am creating WPF application that utilizes a simple navigation menu. When one of the menu items is selected, I want a grid control with buttons to become visible.
I have 3 items on my navigation menu. For purposes of speed and better coding practice, is it better to have 3 seperate grids and they all are visible until the button is clicked and one becomes visible or is it better to have just one whose content gets changed depending on what button is clicked?
I'd prefer to use a container control (e.g. ListBox) instead of a grid. This way you can bind one of your three menus to this container control's DataContext, depending on the button clicked.
If your menu entries are static you could also use a content control and change the ContentTemplate. All three neccessary ContentTemplates can then be defined in XAML.
So i have a datagrid that was built with the mindscape datagrid control that looks like this:
I am not sure if i will be able to do this to the mindscape control, so this will be a question about any multi-column listview or datagrid.
When an item is clicked, and selected, I need to expand the item (hopefully with an animation/storyboard) downwards, to display a bunch of extra information about the item (rating graph, list of available locations, etc).
I honestly have no idea how to begin with this. I'm new to wpf. in winforms i would just create a custom listboxitem that controls it's own height, and listen to the selected event.
A ListView is not the best control to use for your requirements. Instead of the ListView, you should use a DataGrid control. Your required functionality is built in to the DataGrid as it has a row details section... from the DataGrid.RowDetailsTemplate property page on MSDN.
You can customize the data presentation in the DataGrid by adding a row details section. Adding a row details section enables you to group some data in a template that is optionally visible or collapsed.
You define the row details template as either inline XAML or as a resource. A data template that is added as a resource can be used throughout the project without re-creating the template. A data template that is added as inline XAML is only accessible from the control where it is defined.
I want to create a settings menu that looks similar to VLC's advanced settings menu: Treeview on the left and some kind of control collection on the right. The controls on the right should enable the user to manipulate settings that are relevant to the current selection in the tree view. I thought about creating a grid right of the tree view. Then I have a user control for each view that needs to be displayed in the grid, based on the selection
The item in the treeview has a UserControl property that holds a reference to the relevant view. My viewmodel has a SelectedItem property that indicates which item in the tree view is currently selected.
Now I want to bind the content of the grid to the UserControl property of my SelectedItem. But I cannot figure out how to do that. I would prefer to use a XAML based solution instead of clearing the Children property of the grid and adding the user control that I want to display in code each time the SelectedItem property changes.
I'd suggest using ContentControl instead of Grid.
Considering the tree view and the content control are under the same view model: on your view model, add property for selected item (let's call it VMSelectedItem) of same type as the items in the tree view.
In XAML of the tree view add
SelectedItem="{Binding VMSelectedItem}"
In XAML of the content control
Content="{Binding VMSelectedItem.UserControl}"
Now selection in the tree will update the VMSelectedItem property that, in turn, will update the content of the content control.
I suggest you using DataTemplates that you have declared inside your resouces dictionay. You would be using only one instance of each DataTemplate which leaves nice memory footprints. You wouldnt need to store instance of view inside your viewmodel which is the basic idea of mvvm. The viewmodel would completely just holding data and information how you wish the data to be displayed.
As example you have an enum inside your viewmodel with values person, car, tree. Inside your DataTemplateSelector you will have an if on that enum that returns what the desired DataTemplate.
Basically you would have everything central instead of having everything per each TreeViewItem.
I have used a list view that uses a grid view in WPF. I have bound the grid with data and I want to handle right click event on the items of the grid view. Can someone help me out and suggest how to do this??
This question seems to provide what you need:
wpf listview right-click problem